Question

Given the data below, what is the equipment variation as a percent of process variation (Use the ANOVA method)? (Data set in the comments)

Part	Appraiser	Trial	Measurement
1	1	1	8.188
1	1	2	8.129
1	2	1	8.525
1	2	2	8.698
1	3	1	8.591
1	3	2	8.73
1	4	1	9.477
1	4	2	9.262
2	1	1	12.518
2	1	2	12.426
2	2	1	13.399
2	2	2	13.408
2	3	1	14.538
2	3	2	14.372
2	4	1	14.64
2	4	2	14.654
3	1	1	12.633
3	1	2	12.462
3	2	1	13.958
3	2	2	13.963
3	3	1	14.061
3	3	2	13.799
3	4	1	15.382
3	4	2	15.552
4	1	1	18.042
4	1	2	18.139
4	2	1	18.273
4	2	2	18.523
4	3	1	18.687
4	3	2	18.862
4	4	1	20.973
4	4	2	20.808

Answer 1

Two-way ANOVA: Measurement versus Part, Appraiser

Source	DF	SS	MS	F	P-value
Part	3	427.744	142.581	11644.57	0.000
Appraiser	3	21.513	7.171	585.66	0.000
Interaction	9	4.248	0.472	38.54	0.000
Error	16	0.196	0.012
Total	31	453.700

The equipment variation as a percent of process variation=(SS due to Part/SS dur to total)*100=(427.744/453.700)*100=94.28% i.e. 94.28% of total process variation is explained by the equipment variation.

Here we see that Interaction between Part and Appraiser is significantly present (since P-value<0.05) so it is required to test Part and Appraiser individually.